The characteristics of sets are the basis of learning sets well and the key to solving set problems. It mainly refers to the certainty, mutual dissimilarity and disorder of set elements, which provides a basis for us to solve problems, especially the mutual dissimilarity of elements. If we are not careful, we can easily make mistakes.

These three attributes and applications of set elements are explained as follows.

First, pay attention to correctly understand its meaning.

1. certainty:

That is, for any given object, relative to a set, it either belongs to this set or does not belong to this set. The key is to understand the meaning of "determination".

2. Reciprocity:

For a given set, the elements in the set must be different (or different from each other), that is, any two elements in the same set are different objects, and the same object can only be regarded as one element in the set when it is classified into any set.

3. Disorder:

Because the elements in the set are definite and different from each other, and the set with identical elements is an equivalent set, the elements in the set have nothing to do with the order.

Second, pay attention to the correct application of three types of problems.

Example 1 Which of the following propositions is correct?

(1) Very small real numbers can form a set;

⑵ Set {1, 5} and set {5, 1} are different sets;

⑶ Set {( 1, 5)} and Set {(5, 1)} are the same set;

(3)1,∣-∣, 0.5 has five elements.

Analysis: this kind of topic mainly examines the understanding of the concept of set, and the key to solving this kind of problem is to judge according to the certainty, mutual dissimilarity and disorder of elements in the set.

Solution: (1) Very small is a vague concept and there is no clear standard, so it is difficult for us to determine whether an object is in it, which does not conform to the certainty of set elements. So "very small real numbers" cannot form a set, so (1) is wrong.

⑵ {1, 5} is a set consisting of two numbers 1, 5. According to the disorder degree of set elements, it is the same set as {5, 1}, so ⑴ is wrong.

⑶ {( 1, 5)} is a single element set consisting of a point (1, 5). Since (1, 5) and (5, 1) represent two different points, {( 1, 5)}

(4) =, ∣-∣ = 0.5 Therefore, the set consisting of 1, ∣-∣, 0.5 is {1, 0.5}, * * has three elements. Therefore, (4) is also wrong.

Example 2 sets A = {,+,+2}, B = {,}, where A = b, the value of.

Analysis: the most common mistake in this question is to think that the corresponding items of these two sets are the same, list the corresponding relations, and then find the value of, which obviously violates the disorder of the sets.

Solution: ∫a = b, and the disorder of set elements,

∴ There are two situations:

①

Elimination, the solution is = 1, at this time = =, which contradicts the mutual anisotropy of elements in the set, ∴ 1.

(2) Elimination, the solution is =-,or = 1 (truncation), so the value is-.

Comments: This problem uses the disorder of set elements and the element characteristics when two sets are equal to get two equations, which opens the door to solving problems. After calculating these values, the mutual differences of set elements are used to check and ensure the accuracy of the results.

Example 3 Let a = {x ∣+(b+2} x+b+ 1 = 0, br} and find the sum of all elements in A. 。

Wrong solution: (x+1) (x+b+1) = 0from+(b+2) x+b+1.

(1) When b = 0, the sum of elements in x 1 = x2- 1, a is -2.

(2) When b0, x 1+x2 =-b-2.

analyse

The above solution is wrong in (1). When b = 0, the equation has two roots-1 and the set A = {- 1}, so the sum of the elements is-1. The reason for the error is that the "mutual differences" of the elements in the collection are ignored. Therefore, when enumerating collections,

Example 4 evaluates the set {2,3, +4+2}, b = {0 0,7,+4-2,2-}, ab = {3,7}.

Analysis:

∫AB = { 3，7}

∴+4+2 = 7. That is = 1, or =-5.

At this point, many students feel that it is done. In fact, we just found out what the elements in set A and set B are, and whether they meet the mutual anisotropy of elements needs further investigation. When =-5, 2-= 7 appears repeatedly in B, which contradicts the mutual anisotropy of elements, so =-5 should be omitted. When = 1, B = {

∴ = 1

Comments: The certainty, mutual difference and disorder of set elements play an important guiding role in solving problems. Ignoring this, it's a thousand miles away.

Analysis of Error-prone Problems of Sets and Functions and Derivatives

1. When performing the operations of intersection, union and complement of sets, don't forget the special cases of complete sets and empty sets, and don't forget to use the number axis and venn diagram to solve them.

2. Will you solve related problems with the idea of complement set?

3. When solving the solution set of inequality (equation) or definition domain (value domain), is it written in the form of set as required?

Second, set and equation.

Example 3: The range of the real number p is known.

Analysis: Set A is the solution set of equation x2+(p+2)x+ 1=0, so from, we can get two situations:

A=φ, then from, we get:

Equation x2+(p+2)x+ 1=0 has no positive real root. Or (x1x 2 =1>; 0)

therefore

Example 4, known set, set, where x and t are real numbers, find.

Analysis: Set A is the value range of solution set T that makes equation x2+2tx-4t-3=0 φ, and set B is the value range of t that makes equation x2+2tx-2t=0 have a solution, so it is obtained by.

Three. Set and inequality

Example 5: It is known that the set A={a|ax2+4x- 1≥-2x2-a is constant}, and B = {x | x2-(2m+1) x+m (m+1).

If a ∩ b ≠ φ, find the range of m.

Analysis: Set A is the value range of A that makes the inequality ax2+4x- 1≥-2x2-a constant, and set B is the inequality X2-(2m+ 1) X+M (m+ 1).

Solution: The inequality ax2+4x- 1≥-2x2-a holds, and we can get: (a+2) x2+4x+(a- 1) ≥ 0 (★),

(1) When a+2=0, that is, a=-2, the formula (★) can be changed to X >；; 3/4, obviously does not meet the meaning of the question.

(2) When a+2≠0, if formula (★) holds for any x, it must be satisfied.

The solution is A=.

And b = {x | m.

Fourth, setting and solution.

Example 6, if the set is known, find the range of the number A.

Analysis: From the representative element (x, y), these two sets are both point sets, and x2+mx-y+2=0 and x-y+ 1=0 are two equations, so the essence of A∩B≠φ is the intersection of two curves, which we translated into mathematical language: "Parabola x2.

Solution: from, get ①.

Equation ① has at least one real number solution in the interval.

First of all, by.

It was given by x 1+x2 =-(m- 1).

At that time, from x1+x2 =-(m-1) >; 0 and x1x 2 =1>; 0, Equation ① has two positive roots that are reciprocal, so there must be a root in the interval, so Equation ① has at least one root in the interval.

To sum up, the range of m is.

Example 7. If the set is known, find the value of the number A. ..

Solution: (1) When a= 1, the set B = φ, which is consistent with the meaning of the question.

(2) When a≠ 1, it is easy to know that both sets A and B are point sets, in which set A is a point set on a straight line y=(a+ 1)(x-2)+3(x≠2) and set B is a point set on a straight line, so we know that two straight lines are not collinear.

① If two straight lines are parallel,-(a+1) = a+1∴ a =-1.

② If the straight line passes through point (2, 3), the solution is:

To sum up:

Verb (abbreviation for verb) set and derivative

Example 7. As we all know,

A=, then the number of elements in B is

A. there are three B. there are two C. only1D.

Solution: According to the knowledge of derivatives, a = {x | x2-12x+20 ≤ 0} = {x | 2 ≤ x ≤10},

Again, ⅷ

When x∈A, it is easy to know that ∴f(x) is increasing function in the interval [2, 10].

And f (2) < 0, f( 10)>0, the equation f (x) = 0 has one and only one solution in the interval [2, 10].

That is, there is only one element in set B.

Exercise:

Know and ask

Know and ask

Yes, find B.

(4) Known, and find M.

All kinds of mistakes in set learning

Mathematics is a rigorous subject. In collective learning, mistakes will occur unconsciously because of unclear understanding of concepts or incomplete consideration of problems. This paper summarizes all kinds of mistakes in collective learning, hoping to help students avoid such mistakes from happening again.

First, confuse the formation of elements in the set.

Example 1 set, then.

Wrong solution: solving equations

Analysis: The reason for the error is that the form of elements in the set is not clear, which confuses the point set and the number set. The elements in the set are all ordered number pairs, that is, points in the plane rectangular coordinate system, not numbers, so they are point sets, not number sets.

Second, ignore the particularity of empty set.

Example 2 shows that if, the value of is.

Wrong solution: obtained by

By or

Or 3 or

Analysis: Because the particularity of empty set is ignored-empty set is a subset of any set, there is an error of losing the solution. The situation discussed above is only the situation that should be discussed at that time.

The value of is.

Third, ignore the differences between the elements in the set.

Example 3 Known sets, and.

Wrong solution: there must be.

or

Analysis: The feature that the elements in the set should be different from each other is ignored, which leads to the error of increasing the solution. After obtaining the value, we must also check whether the elements in the set should be different from each other.

In fact, (1) did not conform to the characteristics that China elements should be different from each other at that time, so it should be abandoned.

(2) At that time, the elements in,, and the set were different from each other.

The value of is 1.

Fourth, I didn't understand the meaning of the complete work.

Example 4 sets the values of the complete sets,, and.

Wrong solutions: and

or

Analysis: the meaning of the complete works was not correctly understood, which led to the error of adding solutions. The complete set should contain all the elements of the discussion set, so it needs to be tested.

(1) At that time, it was enough at this time.

(2) It should have been abandoned at that time.

Don't understand the essence of things.

Example 5 If, ask whether it is equal.

Wrong solution:

Analysis: I only see the formal differences between the two sets, but I don't understand the essence of things. In fact, they are all even sets, even sets. The two groups should be equal, although in different forms.

In other words,

The elements contained in the two sets are exactly the same.

Six, the misuse of mathematical symbols

Example 6 Use and fill in the blanks

Wrong solution:

The reason for the error is that the difference between the symbols ""and ""is unclear.

""represents the relationship between elements and sets, ""represents the relationship between sets, represents sets, and is also sets.

An example analysis of mathematical thinking method in set

Mathematical thoughts and methods are the soul of mathematics and the bridge from knowledge to ability. The information society requires more and more people to consciously use mathematical ideas and ask and solve problems by mathematical methods. In recent years, the examination of mathematical thinking methods has become a hot issue in the college entrance examination. In order to help students better understand and master the most commonly used basic mathematical thinking methods, the main points of mathematical thinking methods are summarized as follows for development.

First, the idea of equivalent transformation.

When solving the set problem, when a set is not well expressed, it can be transformed into another form first, for example, transforming = B or = A into, into, into, etc.

Example 1 It is known that M ={(x, y) | y = x+a} and N ={(x, y) | x+y = 2}, and the realistic number A makes the range of = hold.

Solution: = Equivalent to the system of equations without solution.

Substitute y = x+a into equation x+y = 2, and eliminate y to get quadratic equation 2x+2ax+a-2 = 0. ①

The problem is transformed into a quadratic equation ① with no real roots, that is, △ = (2a)-4× 2× (a-2) < 0, so a > 2 or a.

Therefore, the range of real number a is {a | a > 2 or a.

Comments: On the basis of understanding the symbol set, we can accurately convert the assembly language into the mathematical problems we learned in junior high school, and then solve the problems with the knowledge and methods we have learned. This transformation can express abstract knowledge with concise and accurate mathematical language and improve the efficiency of solving problems.

Second, the idea of classified discussion

When solving the set problem, we often encounter such a situation: in the process of solving the problem, at a certain step, it is impossible to continue with a unified method and form, because the mathematical object under study already contains many possible situations, so we must choose a standard, divide it into several small problems that can be solved in different forms according to this standard, and solve these small problems one by one, so that the problems can be solved. This is the thinking method of classified discussion.

Example 2 let the set A = {x | x+4x = 0, xR}, b = {x | x+2 (a+1) x+a-1= 0, aR, xR}, if, and find the range of the number a.

Analysis: BA can be divided into three situations: B =, BA and b = a.

Solution: ∫a = {0, -4}, ∴BA is divided into the following three cases:

(1) When B = A, B= {0, -4}, so 0 and -4 are the two roots of the equation X+2 (A+ 1) X+A- 1 = 0. From the relationship between roots and coefficients, we can get:

a = 1 .

Two when the bar, and can be divided into:

① when ①B =, △ = 4 (a+1)-4 (a-1) < 0, and a

② When B≦, B = {0} or b = {-4}, delta = 4 (a+1)-4 (a-1) = 0, the solution is a =- 1, and B = {0} satisfies.

Based on (1) and (2), the value of real number a is a ≤- 1 or a = 1.

Comments: The essence of solving the classified discussion problem is to decompose the whole problem into parts. For planning problems with parameters, it is often necessary to discuss the parameters in categories. Pay attention to "neither weight nor leakage" when classifying. Since an empty set is the proper subset of any non-empty set, it must be the proper subset of the non-empty set, so B =φ also satisfies BA. Therefore, B = and B≠ should be considered in BA, that is to say, the classification discussion of empty set introduction method.

Third, open mind.

Open questions are relative to closed questions with clear conditions and conclusions in middle school textbooks. These questions have a wide range of knowledge, strong comprehensiveness, high requirements for flexible topic selection, novel meaning, ingenious conception, and considerable depth and difficulty. Most of the open questions in the set are open questions with uncertain conclusions.

Example 3 let the set A = {(x, y) | y-x- 1 = 0}, the set B ={(x, y) | 4x+2x-2y+5 = 0}, and the set C ={(x, y) | y = kx+b} exist. If yes, request the values of k and b; If it does not exist, please explain why.

Solution: because, that is, so and.

Substitute y = kx+b into y-x- 1 = 0 to get kx+(2kb-1) x+b-1= 0.

Delta = (2kb-1)-4k (b-1) < 0, that is, 4k-4kb+ 1 < 0. If this inequality has a solution, it should be 16b- 16 > 0, that is, b > 65438.

Substituting y = kx+b into 4x+2x-2y+5 = 0, we get 4x+(2-2k) x+(5-2b) = 0.

Because, △ = (2-2k)-4k (5-2b) < 0, that is, k-2k+8b- 19 < 0. If this inequality has a solution, it should be 4-4 (8b- 19) > 0, and the solution is b < 0. ②.

B = 2。 From inequalities ①, ② and bN.

Substitute b = 2 into the inequality group composed of △ < 0 and △ < 0, and then pay attention to kN, and get K = 1.

So there is a natural number k = 1 and b = 2 that makes.

Comments: In mathematical propositions, it often appears in the form of conclusions "existence (affirmative type)", "non-existence (negative type)" and "existence (discussion type)" which are suitable for a certain condition. "Existence" refers to the existence of an object that is suitable for a certain condition or conforms to a certain property, so as long as one is found by any means, it means existence. "Does not exist" means that it cannot be found by any means. The other is that it does not exist, and the reason needs to be explained.

Eight points for attention in solving problems by set.

When solving the problem of set, if you don't understand the basic concept of set thoroughly and think comprehensively, you will often make mistakes. Therefore, this paper puts forward "eight points for attention" when solving the set problem, hoping to attract students' attention.

1. Pay attention to the differences between the elements in the set.

Any two elements in a set are different, and the same element can only be counted as one element if it is contained in the same set, so there is no element duplication in the set, and ignoring the differences will lead to wrong solutions.

Example 1. Find the value of the number a ..

Misunderstanding: From the meaning of the question:

that is

Analysis: This contradicts the mutual dissimilarity of set elements and is discarded.

2. Pay attention to the meaning of set elements

The elements in the set are meaningful, and a slight negligence will lead to mistakes in solving problems.

Example 2. If, then _ _ _ _.

Wrong solution: obtained by solving the equation:

therefore

Analysis: The reason for the error is that the meaning of the set elements is not correctly understood. The elements in A and B are ordered number pairs, which means points in a plane rectangular coordinate system.

3. Particularity of attention

Is a subset of any set, is the proper subset of any nonempty set, and the union with any set is equal to the set itself. Ignoring its particularity will also cause problems.

Example 3. Given a set, if, find the set c composed of real numbers a. ..

Wrong solution: Because of this.

That's right, so

Analysis: The reason for the error is the lack of information. At that time, the conditions were met, and the following can be obtained:

4. Pay attention to the range of letters.

When parameters are contained in expressions of multiple elements, it is easy to expand the value range of parameters in the operation process, so pay attention to inspection, otherwise there will be wrong solutions.

Example 4. The set is known and

Find the value of a.

Wrong solution: from, know

Analysis: At that time,

The contradiction at this time should be abandoned.

5. Pay attention to the possibility of grading.

Example 5. The range of values of,, and the truth number A is known. ..

Analysis: From the known:

Note: Don't ignore the situation.

6. Pay attention to the importance of classified discussion

Example 6. Given a set, if, find the values of numbers A and B. ..

Analysis: Because, therefore, B contains one or two elements. Through discussion, we can get:

7. Pay attention to implicit conditions

Example 7. Complete works, the value of realistic number a.

Wrong solution: because

Therefore, the conclusion is:

Analysis: the reason for the error is that the implicit condition is not considered, because S is a complete set, so.

When, in line with the meaning of the question;

It didn't fit the meaning of the question, so.

Note: When solving a set problem with parameters, it is necessary to verify whether the results meet the conditions (including implicit conditions) in the problem.

8. Regression definition is also a method.

When you come across a difficult topic, sometimes you will come back to the definition, but it becomes simple.

Example 8. Suppose, then s is ()

Analysis: from the meaning of the question, the sets m and n can be found, and then p, q and r can be found.

Solve by reasoning

pass by

So again by

Example 1, the known set a = {y | y2-(a2+a+1) y+a (a2+1) > 0}, B={y|y2-6y+8≤0}, if a ∩.

Analysis: If this problem is solved directly, the situation is more complicated and it is not easy to get the correct result. It can also be solved if we consider its negative side first and then seek its supplement.

Solution: A = {y | y & gtA2+ 1 or y

Range. draw

By, by

Or.

That is, when a ∩ b = φ, the range of A is or, and when A∩B≠φ, the range of A is obviously its complement, so it is easy to know that the range is.

Commentary: Generally speaking, if the positive situation is complicated when we solve it, we can consider its negative side first, and then use its complement set to get its solution. This is the idea of "complement set".

Example 2: If at least one of the following three equations: x2+4ax-4a+3 = 0, x2+(a- 1) x+a2 = 0, and x2+2ax-2a = 0 has a real root, try to find the range of values of the realistic number A. ..

Analysis: There are seven situations to be considered in the front of this question, and there is only one on the back, that is, "there are no real roots in the three equations". So it's a shortcut to consider its negative side first.

Solution: If all three equations have no real roots, then there are.

. Let A=

So at least one of the three equations has a real root, and the range of real number A is

Example 3: If X, Y and Z are real numbers, prove that at least one of A, B and C is greater than 0.

Analysis: This question directly proves that there are not only many situations, but also it is difficult to find ideas. If we can prove that its negation can't be established, we can be sure that its affirmation can be established.

It is proved that if A, B and C are all less than or equal to 0, then A+B+C is less than or equal to 0.

And a+b+c = x2-2y+y2-2z+z2-2x+π = (x-1) 2+(y-1) 2+(z-1) 2+π-3 > 0 is a constant.